Chapter 17: Additional Aspects of Aqueous Equilibria

Chapter 17:Additional Aspects of Aqueous Equilibria

Section 1:The Common Ion Effect

Objectives

• When you complete this presentation, you will be able to• describe the common ion effect.

Introduction

• Water is the most common and most important solvent on Earth.

• We will be looking in some detail at the application of equilibrium theory and practice to aqueous solutions.• Additional acid-base equilibria• Buffers and acid-base titrations

• Solubility of compounds• Formation of complex ions

The Common-Ion Effect

• We know that sodium salts are strong electrolytes and dissociate completely in aqueous solution.

NaA(aq) → Na+(aq) + A−(aq)

• We also know that certain acids are weak electrolytes and dissociate partially in solution.

HA(aq) ⇄ H+(aq) + A−(aq)


• If we start with a solution of acetic acid, we will set up the equilibrium for the weak acid.

HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)

• If we add sodium acetate to the solution, the additional acetate will drive the equilibrium to the left, decreasing the equilibrium [H+].

• The presence of the added acetate ion causes the acetic acid to ionize less than it normally would.


• Whenever a weak electrolyte and a strong electrolyte contain a common ion, the weak electrolyte ionizes less than it would if it were alone in the system.

• This is called the common-ion effect.

• We can calculate equilibrium concentrations of systems with common ions.


• Sample Exercise 17.1 (pg. 720)

• What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution?

• Plan:• Identify the major species in solution.

• Identify the major sources that affect the concentration of H+.

• Build “i-c-e” table.

• Use K expression to calculate [H+].




• Major species:• HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)




• Major sources that affect the concentration of H+:• HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)




• “i-c-e” table:• HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)initial 0.30 M 0.00 M 0.30 M

change -x +x +x

equilibrium

(0.30 – x) M x M (0.30 + x) M




• K expression:

Ka = 1.8 × 10−5 =

[H+][CH3COO−][HCH3COO]




• K expression:

Ka = 1.8 × 10−5 =

x(0.30 + x)0.30 − x




• K expression:

Ka = 1.8 × 10−5 =

x(0.30 + x)0.30 − x

assume x<<0.30




• K expression:

Ka = 1.8 × 10−5 =

x(0.30)0.30

assume x<<0.30




• K expression:

Ka = 1.8 × 10−5 =

x(0.30)0.30




• K expression:

Ka = 1.8 × 10−5 = = x

x(0.30)0.30




• K expression:

Ka = 1.8 × 10−5 = = x = [H+]

x(0.30)0.30




• calculate pH:

pH = −log[H+]




• calculate pH:

pH = −log[H+] = −log(1.8 × 10−5)




• calculate pH:

pH = −log[H+] = −log(1.8 × 10−5) = 4.74



• Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

• Plan:• Identify the major species in solution.

• Identify the major sources that affect the concentration of H+ & F−.

• Build “i-c-e” table.

• Use K expression to calculate [H+] & [F−].




• Major species:

HF(aq) ⇄ H+(aq) + F−(aq)




• Major sources that affect [H+]:





• “i-c-e” table:


initial 0.20 M 0.10 M 0.00 M

change -x +x +x

equilibrium

(0.20 – x) M

(0.10 + x) M

x M




• K expression:

Ka = 6.8 × 10−4 =[H+][F−]

[HF]




• K expression:

Ka = 6.8× 10−4 =(0.10 +x)x0.20 − x




• K expression:

Ka = 6.8× 10−4 =(0.10 +x)x0.20 − x

assume x<<0.30




• K expression:

Ka = 6.8× 10−4 =(0.10)x

0.20

assume x<<0.30




• K expression:

Ka = 6.8× 10−4 =

x = (6.8× 10−4)

(0.10)x0.20

0.200.10




• K expression:

Ka = 6.8× 10−4 =

x = (6.8× 10−4) = [F−]

(0.10)x0.20

0.200.10




• K expression:

Ka = 6.8× 10−4 =

x = (6.8× 10−4) = [F−] = 1.4 × 10−3 M

(0.10)x0.20

0.200.10




• Find pH:

[H+] = 0.10 M − 1.4 × 10−3




• Find pH:

[H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M




• Find pH:

[H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M

pH = −log[H+]




• Find pH:

[H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M

pH = −log[H+] = −log(0.10)




• Find pH:

[H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M

pH = −log[H+] = −log(0.10) = 1.00


• Sample Exercises 17.1 and 17.2 both involve weak acids.

• We can also use the same techniques with weak bases.

• For example, adding NH4Cl to an aqueous solution of NH3 will cause the NH3 to dissociate less and lower the pH.


• The techniques of Sample Exercises 17.1 and 17.2 may be used to solve homework problems 17.15 and 17.17.

Chapter 17: Additional Aspects of Aqueous Equilibria

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